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IF 11 California Test 301 1978 equivalent factor is given for concrete. Guncrally, (Equation #7) the untreated materials under 1)ortland cement con- Ct— Cf'/T MIC pavement extend under the shoulder area. (Lquation #8) Therefore, when testing materials which underlie Portland cement concrete pavements,the traffic in. GE = 0.0032 (7%) (100 — 1� Ilex and gravel equivalent factors of the applicable For the example: layers of the shoulder structural section should be GL':= 0.0032 (8,0) (100— 10) = 2.30' used for calculating R-value by expansion pressure. The brawl equivalent may also be determined by If the values used for TI and Gr in the calculation the nomograph on Figure 19. of the R-s-alue by expansion pressure are modified later, the Ibvalce must be recalculated. D. SURFACE THICKNESS Traffic infOrinstion needed for the design of any. To determine the thickness of asphalt concrete re- State High-av is ierailablc from the 'frafiic Depart_ quired use the design equation (5) or nomograph in rnent or from th^ appropriate District Traffic Dc- Figure 19.Assume it base material with a 78 R-value. partment. With a straightedge, intersect Scale E at 78 R-value E. REPORTING RESULTS and Scale F at 8.0 traffic index. The intersection of this line with Scale G is the thickness of cover, in All Test data used to determine the suitability of terms of gravel equivalent,required over the:78 R- the material for roadbed construction along with the value base material. Using this value (0.56' in the design determinations of thickness of corer are re- example) of gravel equivalent as a turning point, ported to the various agencies concerned on Form intersect Scale H at the appropriate value of Gr for T.L.-375, The reverse side of this form contains an the AC.This value of Gr is found from the following explanation of the individual items on the form. equation: PART VIII. METHOD OF CALCULATING THE (Equation #9) DESIGN THICKNESS OF PAVEMENT 2.5 (5.141 TI)0' < 2.3 SECTIONS BASED ON THE For convenience,the gravel equivalent factors are RESULTS OF THE R-VALUE tabulated in Table 2.For the example,from Table 2, TESTS Gr=2.01.The intersection of this line with Scale I AM A. SCOPE gives 0.25'of AC required. Use 0.30'of AC. This part of the procedure describes the method E. BASE THICKNESS for designing a pavement section when the R-values Using a 50 11-value for subbase material,Figure 19 Of the base_, subbase, and basement soil are known. indicates a gravel equivalent of 1.28'needed over the An example calculation i. used to demonstrate this subbase material.Since the 0.30'AC is equivalent to procedure. 0.60'of gravel,.tib'of gravel remains to be satisfied by PROBLEM:Design a structural section for a road the base material.A C,of 1.1 for a good crushed rock carrying the truck traffic indicated in Part VII-C-2a. product would indicate 0.62' to be satisfactory. Use AM (TI 8.0).The hose has-in R-value of 78,the subbase 0.65'of base material. 50,and the basement soil 10. The gravel equivalent factors are given in Table 2 in Part VII-C-2b. F. SUBBASE THICKNESS DESIGN B. TRAFFIC INDEX CALCULATION soil was found in (H.) above,a 1011-value basement soil with TI =8.0 requires a gravel equivalent of The traffic index is calculated as in Part VII-C-2a 2.30'.Summing up the section thus far: of this test method. 0.30'of surface=0.60' GE C. PAVEMENT THICKNESS CALCULATION 0.65'of base =0.72' GE The required total thickness, in terms of gravel Total................ 1.32'GE equivalent (GE) is determined by equation (8) Required thickness of subbase is, therefore: which is derived from equation (5) when: 2.30 -- 1.32= .98' Use 1.00'subbase. I 28